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I am reading the following question:

Inequality between Frobenius and nuclear norm

I know $$\|X\|_* = \sum_{i=1}^r \sigma_i(X)$$and $$\|X\|_F = \bigg(\sum_{i=1}^r \sigma^2_i(X)\bigg)^{1/2}$$

I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.

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Cauchy—Schwarz tells you that $$ \sum_{i=1}^r a_i b_i \leq \left(\sum_{i=1}^r a_i^2\right)^{1/2}\left(\sum_{i=1}^r b_i^2\right)^{1/2} $$ Apply it with $a_i = 1$, $b_i = \sigma_i(X)$ (for $1\leq i\leq r$).

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